Q: about gravity...Let's say, you could dig a 1' wide hollow tunnel from the north pole to the south pole...

Question

without dealing with temperature issues, gasses, lava, etc. (Remember, it is a question about gravity) If you were able to drop a golf ball down the hole, wouldn't the gravitational pull force it to speed to the center of the earth then pass the center, slow up, eventually return to the center,and float in the center point, or what? Does my description make sense and what do you think would happen? I'm no rocket scientist but this is a question I've been pondering for awhile.

Answer 1

This is a very interesting question, with an even more fascinating answer, as I think you'll agree. Here is an answer I first prepared for a question about boring a hole through the centre from ANY point on the surface. I first considered a non-rotating Earth, to get rid of that small complication. But in YOUR case, from north pole to south pole, what I then wrote would be valid independently of the Earth's rotation, since the motion is along its spin axis. In that case the question of the apparent "centrifugal" or "coriolis" "forces" would not then arise :

If the hole were friction free, you'd continuously accelerate downwards, passing through its centre with your highest velocity, then decelerate until you reached your own personal antipodal point (the point exactly on the other side of the centre). Unless you were then able to step out of the hole, you'd start back again, doing exactly what you did before. In other words, you'd oscillate back and forth like that for eternity.

The time each trip to the other side of the Earth would take depends upon the details of how the Earth's density (assumed spherically symmetric) is radially distributed. A precise answer can be given if one assumes the Earth to be a UNIFORM SPHERE, with its density everywhere equal to its mean density of ~ 5.5 grams per cubic centimetre. Each one-way trip would then consist of one "one-way swing" of perfect simple harmonic motion (SHM), and take ~ 42 1/4 minutes. The complete SHM, round-way trip would take ~ 84 1/2 minutes.

It is NO COINCIDENCE that this latter time is the circular orbital period of the LOWEST POSSIBLE Earth-orbiting satellite. (For years my better students have been able to show why this is so --- even under examination conditions.)

What is even more intriguing but much less well known is this:

THE HOLE DOESN'T HAVE TO BE DRILLED THROUGH THE CENTRE !!

That's right --- idealized friction-free travel could be performed to ANYWHERE on Earth's surface, just under gravity, in a time of order 3/4 of an hour, with round-trip times of about an hour and a half. (My very best students have been able to prove this, also.) There's only be about enough time to eat your lunch during this ultimate "free lunch" --- or should we say "free launch?" !!

Live long and prosper.

P.S. By the way, more scientific facts:

(1.) As Isaac Newton himself showed, there is NO gravitational force whatsoever at the very centre of ANY spherically symmetrical distribution of matter. So it's completely wrong to assert that "gravity" (per se) would "crush you."

People often get confused about the separate consequences of gravity and pressure. The PRESSURE in the surrounding material there (arising from the total weight of the overlying material) would certainly CRUSH YOU, but ONLY if you were encased in that material of course. But since you're falling through a HOLE --- without that direct embedding --- that's of no concern to you. This leaves aside the great engineering challenge of building and maintaining the walls of the hole under that same pressure!

(2.) If the Earth were of UNIFORM DENSITY, the gravitational acceleration would indeed reduce linearly with your distance from the centre. It's that LINEAR dependence that is responsible for the motion being SIMPLE HARMONIC MOTION, in that case, and in that kind of case only.

Answer 2

You're right. The ball would fall to the centre of the earth. However, once it got there, it would oscillate north to south for quite some time, and theoretically it would eventually settle directly on the earth's centre of gravity.

If you wanted to take the sun into configuration, the ball would actually move around the earth's centre of gravity, due to the force of the sun's gravity on the ball, in addition to the earth's. It would be a very small force, but it would keep the ball in circular motion (not rotational) as the earth orbits the sun. The moon would also have a small effect on the ball's motion, as well as any other massive objects in the area.

Answer 3

you are correct, lets assume you drop from north pole. first the ball will be attracted towards the center by earth's gravity then after that ball cotinues to move towards south pole because of inertia of motion and eventually as the acceleration due to gravity acting against the motion, it will stop and continue towards the north pole. this is of course neglecting air resistance and possibe loss of momentum by collision at the walls and disregarding other issues like larva, gases, instability of core of earth etc. yes one way motion will take about 42 mins if no force will be applied while dropping the ball. Theoritically the ball continues to move in the same ocillating motion without losing any component of motion and this is basically SIMPLE HARMONIC MOTION. practically there will be air resistance and very high chance of collision with the walls and eventually the ball will rest on the centre of the the earth.
This explanation is regarding Newton's Universal Law of Gravitation. There will not be significant affect due to sun or moon's gravity.
I am not sure how Einstein's general theory of relativity defines it but so far as my knowledge is concerned, he defined gravity as distortion in spacetime curvature which means the ball will simply fall and rest on the centre of earth as it possibly gives more deformation on the space time curvature because the centre is supposedly massive because of the hevay metal present there.

Answer 4

You've got it essentially correct. If you could eliminate air resistance and any rubbing against the walls, you could drop the ball in and it would go back and forth from pole to pole forever.

But with air resistance, the motion will be less and less (like when you bounce a ball) and it will eventually end up in the center where it's being pulled equally by the Earth's gravity in all directions.

By the way, the one way trip would take 42 minutes.

Answer 5

Ossilated inside the hole

Answer 6

Yeah, you'd be crused by gravity before you'd even get half way to the center.